# For #f(t)= (t^3-t+1,t^2-t)# what is the distance between #f(2)# and #f(5)#?

The distance between

Hence distance between the two is

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To find the distance between ( f(2) ) and ( f(5) ), we first need to calculate the values of ( f(2) ) and ( f(5) ), then find the distance between these two points using the distance formula in two dimensions.

Given ( f(t) = (t^3 - t + 1, t^2 - t) ), we can find ( f(2) ) and ( f(5) ) as follows:

( f(2) = (2^3 - 2 + 1, 2^2 - 2) = (7, 2) )

( f(5) = (5^3 - 5 + 1, 5^2 - 5) = (121, 20) )

Now, using the distance formula for two dimensions:

( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} )

Where ( (x_1, y_1) ) are the coordinates of ( f(2) ) and ( (x_2, y_2) ) are the coordinates of ( f(5) ).

Substituting the values:

( \text{Distance} = \sqrt{(121 - 7)^2 + (20 - 2)^2} )

( = \sqrt{(114)^2 + (18)^2} )

( = \sqrt{12996 + 324} )

( = \sqrt{13320} )

( \approx 115.45 )

So, the distance between ( f(2) ) and ( f(5) ) is approximately ( 115.45 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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